Give the Lewis Structure for the Polyatomic Ion SO4²⁻

The Lewis structure for the sulfate ion (SO₄²⁻) can be constructed by following these steps:

1. Count the total number of valence electrons:
   Sulfur (S) has 6 valence electrons, and each oxygen (O) has 6 valence electrons. Since there are four oxygen atoms, that contributes 4 × 6 = 24 electrons. The sulfate ion has a 2− charge, which means we add 2 extra electrons. So, the total is:
   6 (S) + 24 (4 × O) + 2 (charge) = 32 valence electrons.
2. Determine the central atom:
   Sulfur is less electronegative than oxygen, making it the central atom.
3. Draw single bonds:
   Connect the sulfur atom to each of the four oxygen atoms with single bonds. This uses up 8 electrons (4 bonds × 2 electrons per bond).
4. Distribute the remaining electrons:
   After forming the bonds, we have 32 − 8 = 24 electrons left. Distribute these among the oxygen atoms to satisfy the octet rule. Place three lone pairs (6 electrons) on each oxygen atom, leaving one oxygen with only one lone pair (2 electrons) and participating in a double bond with sulfur; this means two of the oxygen atoms will have a complete octet (8 electrons). The last oxygen with the double bond also has 8 electrons.
5. Check formal charges:
   The formal charges should be minimized. In this structure, sulfur has a formal charge of 0, while three of the oxygens with single bonds have a formal charge of -1, and the oxygen with the double bond has a formal charge of 0. This accounts for the overall charge of -2 for the sulfate ion.

The final Lewis structure can be represented as follows:

S
/ \/
O O
\ /
O

In this structure, the sulfur atom is in the center, bonded to four oxygen atoms, with two double bonds and two single bonds with corresponding lone pairs on the oxygen atoms. This visualization respects the octet rule and accounts for the overall charge of the ion.